Question

The roots of a Quadratic Equation are 3+root2 and 3-root 2 then find that equation.
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Answer 1

▪Given :-

For a Quadratic Equation Zeros are

$\bf \alpha = 3 + \sqrt{2} \\ \bf \beta = 3 - \sqrt{2}$

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▪To Find :-

The Quadratic Equation .

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▪Key Point :-

If sum and product of zeros of any quadratic Equation are s and p respectively,

Then,

The quadratic Equation is given by :-

$\bf  {x}^{2}  - s \: x + p=0$

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▪Solution :-

Here,

$\alpha = 3 + \sqrt{2} \\ \beta = 3 - \sqrt{2}$

Sum of Zeros = s = 6

and

Product of Zeros = p = 7

So,

Required Equation should be

$\bf{x}^{2}  - 6x +  7=0$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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▪Verification :-

$\bf {x}^{2} - 6x + 7 = 0 \\ \\ \sf x = \dfrac{6 \pm \sqrt{36 - 28} }{2} \\ \\ \sf x = \dfrac{6 \pm \sqrt{8} }{2} \\ \\ \sf x = \frac{6 \pm2 \sqrt{2} }{2} \\ \\ \implies \bf x = 3 \pm \sqrt{2}$

So Zeros are

$\bf3 + \sqrt{2} \: \: and \: \: 3 - \sqrt{2}$

Hence Verified

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